Question

22. If A.B.C are angles of a triangle and none of them is equal to 2. then prove that
tanA + tanB + tanc = tan A. tanB. tanC.
$561x3 = 30 \div 3 \sqrt{ = = 0 \sqrt[2516 | \leqslant > \leqslant \geqslant 1| ]{?} }$
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Answer 1

Answer:

We have to prove that tan A + tan B + tan C = tan A*tan B*tan C for any non-right angle triangle.

For any triangle the sum of the angles is equal to 180 degrees. If we take a triangle ABC, A + B + C = 180 degrees.

A + B + C = 180 or A + B = 180 - C

tan (A + B) = tan (180 - C) = tan C

=> (tan A + tan B)/(1 - tan A*tan B) = tan C

=> tan A + tan B = tan C - tan A*tan B*tan C

=> tan A + tan B + tan C = tan A*tan B*tan C

This proves that tan A + tan B + tan C = tan A*tan B*tan C